Blocks A (mass 4.00 kg ) and B (mass 6.00 kg ) move on a frictionless, horizonta
ID: 1460272 • Letter: B
Question
Blocks A (mass 4.00 kg ) and B (mass 6.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 5.00 m/s . The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of block Find the maximum energy stored in the spring bumpers. Find the velocity of block A when the energy stored in the spring bumpers is maximum. Find the velocity of block B when the energy stored in the spring bumpers is maximum. Find the velocity of block A after they have moved apart. Find the velocity of B after they have moved apart.
Explanation / Answer
mass of the bock A is, m1=4 kg
mass of the bock B is, m2=6 kg
velocity of the block A is u1=5 m/sec
velocity of the block B is u2=0
a)
by using law of conservation of energy,
m1*u1+m2*u2=(m1+m2)*v
4*5+0=(4+6)*v
final velocity v=2 m/sec ( when the spring gets compression)
by using law of conservatio of enegry,
maximum energy stored U=K1-K2
U=1/2*m1*u1^2-1/2*(m1+m2)v^2
U=1/2*4*5^2-1/2*(4+6)*2^2
U=30 J
b)
if energy stored in the bumper is maximum,
velocity of block A is, v=2 m/sec
velocity of block B is, v=2 m/sec
c)
after moving back,
velocity of the block A is,
v1=((m1-m2)/(m1+m2))*u1
v1=((4-6)/(4+6))*5
v1=-1 m/sec
and
velocity of the block B is,
v2=((2*m1)/(m1+m2))*u1
v2=((2*4)/(4+6))*5
v2=4 m/sec